Question

The number of flaws in a fiber optic cable follows a Poisson distribution. It is known that the mean number of flaws in 50m of cable is 1.2. a) What is the probability of exactly three flaws in 150m of cable? b) What is the probability of at least two flaws in 100m of cable? c) What is the probability of exactly one flaw in the first 50m of cable, and exactly one flaw in the second 50m of cable?

Answer 1

Answer:

(a) The probability of exactly three flaws in 150 m of cable is 0.21246

(b) The probability of at least two flaws in 100m of cable is 0.69155

(c) The probability of exactly one flaw in the first 50 m of cable, and exactly one flaw in the second 50 m of cable is 0.13063

Step-by-step explanation:

A random variable X has a Poisson distribution and it is referred to as Poisson random variable if and only if its probability distribution is given by

$p(x;\lambda)=\frac{\lambda e^{-\lambda}}{x!}$ for x = 0, 1, 2, ...

where $\lambda$, the mean number of successes.

(a) To find the probability of exactly three flaws in 150 m of cable, we first need to find the mean number of flaws in 150 m, we know that the mean number of flaws in 50 m of cable is 1.2, so the mean number of flaws in 150 m of cable is $1.2 \cdot 3 =3.6$

The probability of exactly three flaws in 150 m of cable is

$P(X=3)=p(3;3.6)=\frac{3.6^3e^{-3.6}}{3!} \approx 0.21246$

(b) The probability of at least two flaws in 100m of cable is,

we know that the mean number of flaws in 50 m of cable is 1.2, so the mean number of flaws in 100 m of cable is $1.2 \cdot 2 =2.4$

$P(X\geq 2)=1-P(X<2)\\P(X\geq 2)=1-(P(X=0)+P(X=1))$

$P(X\geq 2)=1-p(0;2.4)-p(1;2.4)\\\\P(X\geq 2)=1-\frac{2.4^0e^{-2.4}}{0!}-\frac{2.4^1e^{-2.4}}{1!}\\\\P(X\geq 2)\approx 0.69155$

(c) The probability of exactly one flaw in the first 50 m of cable, and exactly one flaw in the second 50 m of cable is

$P(X=1)=p(1;1.2)=\frac{1.2^1e^{-1.2}}{1!}\\P(X=1)\approx 0.36143$

The occurrence of flaws in the first and second 50 m of cable are independent events. Therefore the probability of exactly one flaw in the first 50 m and exactly one flaw in the second 50 m is

$(0.36143)(0.36143) = 0.13063$

Answer 2

Using the Poisson distribution, it is found that there is a :

a) 0.2125 = 21.25% probability of exactly three flaws in 150m of cable.

b) 0.3084 = 30.84% probability of at least two flaws in 100m of cable.

c) 0.1306 = 13.06% probability of exactly one flaw in the first 50m of cable, and exactly one flaw in the second 50m of cable.

In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by:

$P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}$

The parameters are:

• x is the number of successes
• e = 2.71828 is the Euler number
• $\mu$ is the mean in the given interval.

Item a:

Mean number of flaws of 1.2 per 50m, thus, for 150m, $\mu = 1.2\frac{150}{50} = 3.6$

The probability is P(X = 3), then:

$P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}$

$P(X = 3) = \frac{e^{-3.6}(3.6)^{3}}{(3)!} = 0.2125$

0.2125 = 21.25% probability of exactly three flaws in 150m of cable.

Item b:

Mean number of flaws of 1.2 per 50m, thus, for 100m, $\mu = 1.2\frac{100}{50} = 2.4$

The probability is:

$P(X \geq 2) = 1 - P(X < 2)$

In which

$P(X < 2) = P(X = 0) + P(X = 1)$

Then

$P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}$

$P(X = 0) = \frac{e^{-2.4}(2.4)^{0}}{(0)!} = 0.0907$

$P(X = 1) = \frac{e^{-2.4}(2.4)^{1}}{(1)!} = 0.2177$

Then

$P(X < 2) = P(X = 0) + P(X = 1) = 0.0907 + 0.2177 = 0.3084$

0.3084 = 30.84% probability of at least two flaws in 100m of cable.

Item c:

50m, thus, $\mu = 1.2$

For each 50m, the probability of exactly one flaw is P(X = 1):

$P(X = x) = \frac{e^{-\mu}\mu^{x}}{(x)!}$

$P(X = 1) = \frac{e^{-1.2}(1.2)^{1}}{(1)!} = 0.3614$

The intervals are independent, thus:

$p = 0.3614^2 = 0.1306$

0.1306 = 13.06% probability of exactly one flaw in the first 50m of cable, and exactly one flaw in the second 50m of cable.

A similar problem is given at https://brainly.com/question/16912674